Question

prove for each integer n, n is even if and only if 4 divides n^2

Answer 1

hint
a general even number looks like \(n=2j\) for some \(j\)

Answer 2

I am having trouble....

Answer 3

better yet what I am having trouble is proving the contrapositive of: if n is odd then 4 does not divide n^2.

Answer 4

Just like Satellite wrote the general form of an even number, what is the general form of an odd number? What happens when you square it?

Answer 5

I got the first part which is: for each integer n, if n is even, then 4 divides n^2.

Answer 6

nerdguy2535, the def of an odd number is 2k+1, I already worked it out which (2k+1)^2=4k^2+4k+1, then I factored it

Answer 7

and got 2(2k^2+2k)+1, therefore n^2=2q+1 for some q is element of integers

Answer 8

Nice. Is it possible for an odd number to be divisible by 4?

Answer 9

no, I guess what I am not seeing is how to make the connection in the end in words, to finalize my proof...

Answer 10

why go that route? you can do it directly much faster and mostly direct proofs are nicer and more convincing

Answer 11

takes only one or two lines in each direction

Answer 12

He's already done doing the contrapositive though.

Answer 13

satellite73, I tried proving it directly but it got messy with the square root, because original section of proof is: if 4 divides n^2, then n is even. Unless I was doing it wrong?

Answer 14

for example if \(n\) is even, then \(n=2j\) for some integer \(j\) making
\[n^2=(2j)^2=4j^2\] divisible by \(\) done

Answer 15

ooh i see

Answer 16

there is always the fundamental theorem of arithmetic
but i guess the contrapositive works nicely without it

Answer 17

what is the fundamental theorem of arithmetic?